Answered

Get the answers you've been looking for with the help of IDNLearn.com's expert community. Find reliable solutions to your questions quickly and easily with help from our experienced experts.

Assuming complete dissociation of the solute, how many grams of kno3 must be added to 275 ml of water to produce a solution that freezes at − 14. 5 ∘c ? the freezing point for pure water is 0. 0 ∘c and kf is equal to 1. 86 ∘c kg mol−1.

Sagot :

The mass of KNO₃ that is required to be added to 275 ml of water is; 108.3 g

Dissolution of a Solute

We are given;

K f =  1.86 °C.kg/m

Mass of solution = 275 g = 0.275 Kg

Freezing point of solution = −14.5 °C

Freezing point of pure water = 0.0 °C

We know that;

Δ T = Freezing point of solve nt - Freezing point of solut ion

Thus;

Δ T =  0°C - ( -14.5 °C)

Δ T = 14.5 °C

Also;

Δ T = K f * m * i

where;

K f is free zing con stant

m is m o l a l i t y

i is V a n t H o f f factor

Since KNO₃ produces two particles then; i = 2

Making the m the subject of the formula gives;

m = ΔT /(K × i)

m =  14.5 °C/(1.86 °C . kg/m o l × 2)

m = 3.898 m

Now;

M o l a l i t y = number of mo les/mass of solution in kilograms

Thus;

number of mo les of solute = m o l a l i t y × mass of solution in kilograms

number of mo les of solute= 3.898 m × 0.275 Kg

number of mo les of solute= 1.072 moles

Thus;

Number of mo les = mass/mo lar mass

M o lar mass of KNO₃ = 101 g/mol

Mass = Number of mo les ×  101 g/mol

Mass = 1.072 moles ×  101 g/mol

Mass of KNO₃ = 108.3 g

Read more about dissolution of a solute at; https://brainly.com/question/1945924

Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.